Let $f(x)$ be twice differentiable on $[a, +\infty)$ and $\lim_{n\rightarrow\infty} f(n)$ exist. If $\lim_{x\rightarrow+\infty} f''(x)$ exists, prove that $\lim_{x\rightarrow+\infty} f(x)$ exists and $$\lim_{x\rightarrow+\infty} f'(x)=\lim_{x\rightarrow+\infty} f''(x)=0$$
According to the answer here, we know that
If $\lim_{x\rightarrow\infty} f(x)$ exists and $f''$ is bounded, then
$\lim_{x\rightarrow\infty} f'(x)=0$.
However, in this problem, the limit of $f$ is along the sequence of positive integers. How can we prove this?
Any assistance or feedback regarding this issue would be highly appreciated.
Best Answer
If $f''(x) \to L$ then by applying L'Hospital's Rule (form "$\text{anything} /\infty$") we get $f'(x) /x\to L$. Applying the rule once again we get $f(x) /(x^2/2)\to L$. This means that $f(n) /n^2\to L/2$. And since $f(n) $ tends to a limit as $n\to\infty$ we have $f(n) /n^2\to 0$ so that $L=0$.
Next note that since $f(n) $ tends to a limit it follows that there is a sequence $c_n\in(n, n+1)$ such that $f'(c_n) \to 0$. This is precisely given by mean value theorem as $f(n+1)-f(n)=f'(c_n)$.
Next we note that $$f'(x) =f'(c_{[x]} ) +(x-c_{[x]}) f''(d_x) $$ where $[x] $ is integer part of $x$ and $d_x$ is between $c_{[x]} $ and $x$. Now as $x\to\infty$ the first term tends to $0$ and the second term is a product of two factors: $f''(d_x) $ which tends to $0$ and $(x-c_{[x]}) $ which is bounded in absolute value by $1$. Thus second term also tends to $0$. Thus $f' (x) \to 0$.
Now use $$f(x) =f([x]) +(x-[x]) f'(\xi) $$ to conclude that $f(x) $ tends to the same limit as that of $f(n) $.